Question

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol of I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium? H2+I2<-> 2HI

Answer 1

Your answer is: Kc = [HI]² / {[H2] [I2]}
53.3 = (2x)² / {(0.400M - x)(0.400M - x)}
sqrt(53.3) = 2x / (0.400M - x)
2.92 - 7.30x = 2x
x = 0.314M

[HI] = 2x = 0.628M

:) hopefully this helped. (: